Question

我有一个称为“游戏”的模型，其中有许多由模型帖子连接的帖子。当我按发布的日期对游戏进行排序时，它会返回来自模型游戏的重复项。我想知道如何在最新发布日期之前显示游戏，而不会返回重复项。

Views.py

<script src="https://cdnjs.cloudflare.com/ajax/libs/jquery/3.3.1/jquery.min.js"></script>
<a href="#">Just a pound (should not log anything to the console)</a><br>
<a href="#more">More than a pound (should log "heavy" to the console, and take you to Heavy)</a><br>
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<p id="more">Heavy</p>
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Models.py

from django.shortcuts import render, get_object_or_404
from library.models import Game
from .models import Post
from django.views.generic import (
ListView,
DetailView
)

# Create your views here.
def home(request):

    context = {
        'recent': Game.objects.all().order_by('-post__date_published')[:5],
        'posts': Post.objects.all(),
}
    return render(request, 'main/home.html', context)

class TitlePostListView(ListView):
    model = Post
    template_name = 'main/title_posts.html'
    context_object_name = 'posts'
    paginate_by = 5


    def get_queryset(self):
        title = get_object_or_404(Game, title=self.kwargs.get('title'))
        return Post.objects.filter(game=title).order_by('-date_published')

    def get_context_data(self, **kwargs):
        context = super(TitlePostListView, self).get_context_data(**kwargs)
        context['game'] = get_object_or_404(Game, title=self.kwargs.get('title'))
        return context

class PostDetailView(DetailView):
     model = Post

Answer 1

您在这里寻找的是distinct()。

Game.objects.all().order_by('-post_set__date_published').distinct()[:5]

我也看不到您将Post和Game连接的位置，因此我假设您正在使用尚未发布的其他代码进行操作，并且当前使用的{{ 1}}在工作。

按模型B对模型A排序，返回重复项

1 个答案: